anonymous
  • anonymous
Find the limit to 0 of.... [(e^2x)-1]/x Is it possible to do this w/out a calculator?
Mathematics
chestercat
  • chestercat
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zzr0ck3r
  • zzr0ck3r
do you know l`hopital rule yet?
anonymous
  • anonymous
no... :(
anonymous
  • anonymous
You can't have a zero in the denominater. The answer is 2

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zzr0ck3r
  • zzr0ck3r
dont just give answer, help them find answer....I dont know how to solve this wthout lhopitals rule
anonymous
  • anonymous
@zzr0ck3r yeah, I'm trying to find a direct way to do this, without l'Hopital's
anonymous
  • anonymous
idk, I'm just practicing for a test and this question was online...
zzr0ck3r
  • zzr0ck3r
you nee to know how to tke a derivative to do this, you will lear that soon
anonymous
  • anonymous
I know how to take a derivative, but how will that help me solve it?
zzr0ck3r
  • zzr0ck3r
take the derivative of the num and the the denom. then run the limit
zzr0ck3r
  • zzr0ck3r
you will have (2e^2x)/1 now run the limi
zzr0ck3r
  • zzr0ck3r
limit*
anonymous
  • anonymous
Oh, I've got one: Well we know: \[ e^x-1\sim x\\ x\to 0 \] So: \[ \lim_{x\to 0}\frac{e^{2x}-1}{x}=\lim_{u\to 0}\frac{e^{u}-1}{\frac{1}{2}u} \]Where \(u=2x\) Due to that: \[ \lim_{x\to 0}\frac{u}{\frac{1}{2}u}=\frac{1}{\frac{1}{2}}=2 \]

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